The mean and variance of a random variable X | vedclass

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(A) For a binomial distribution,the mean is given by $np = 4$ and the variance is given by $npq = 2$.Dividing the variance by the mean,we get $\frac{n

Vedclass · Accepted Answer

$1/32$

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(A) For a binomial distribution,the mean is given by $np = 4$ and the variance is given by $npq = 2$.Dividing the variance by the mean,we get $\frac{npq}{np} = \frac{2}{4}$,which simplifies to $q = \frac{1}{2}$.Since $p + q = 1$,we have $p = 1 - \frac{1}{2} = \frac{1}{2}$.Substituting $p = \frac{1}{2}$ into $np = 4$,we get $n(\frac{1}{2}) = 4$,so $n = 8$.The probability mass function for a binomial distribution is $P(X = k) = \binom{n}{k} p^k q^{n-k}$.For $X = 1$,we have $P(X = 1) = \binom{8}{1} (\frac{1}{2})^1 (\frac{1}{2})^{8-1} = 8 	imes (\frac{1}{2})^8 = 8 	imes \frac{1}{256} = \frac{1}{32}$.

The mean and variance of a random variable $X$ having a binomial distribution are $4$ and $2$ respectively,then $P(X = 1)$ is

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